Reaction Mechanisms — Stoichiometry and Kinetic Order

By Prof. Marco Ruzzi.

Theoretical background

A reaction mechanism consists of elementary steps. A formal decomposition of an overall equation into balanced subreactions does not by itself identify those steps. Molecularity counts the reacting molecular entities in an elementary event; it is not the number of different chemical species in a balanced overall equation.

For an elementary step, its reaction order is consistent with its molecularity. A first-order rate coefficient has units s⁻¹.

Exercise 9 — Acetylene to benzene

The conversion of acetylene to benzene can be written as:

\[\mathrm{C}_2\mathrm{H}_2(g)\longrightarrow\tfrac13\mathrm{C}_6\mathrm{H}_6(l).\]

It is the formal sum of the following two subreactions:

(I)

\[\begin{aligned} \mathrm{C}_2\mathrm{H}_2(g)+\tfrac52\mathrm{O}_2(g)\\ \longrightarrow2\mathrm{CO}_2(g)+\mathrm{H}_2\mathrm{O}(l). \end{aligned}\]

(II)

\[\begin{aligned} 2\mathrm{CO}_2(g)+\mathrm{H}_2\mathrm{O}(l)\\ \longrightarrow\tfrac13\mathrm{C}_6\mathrm{H}_6(l)+\tfrac52\mathrm{O}_2(g). \end{aligned}\]

Kinetic measurements give k₁=4.8×10⁻⁵ s⁻¹ and k₂=7.1×10⁻⁶ s⁻¹ for (I) and (II), respectively. Determine whether these subreactions constitute a mechanism of elementary steps for the overall reaction.

Solution

To constitute a mechanism, the subreactions must be elementary events. These equations instead express a formal stoichiometric decomposition. Their fractional molecular coefficients cannot describe individual elementary events as written.

The given constants have units s⁻¹, consistent with first-order kinetics. They do not turn the displayed subreactions into elementary steps. In particular, molecularity cannot be assigned as “two” merely because each subreaction contains two different reactant species. The distinction is between an overall stoichiometric identity and the molecular events of a reaction mechanism.